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A Capacitor stores 20.0 of energy when charged to 60.0v. Calculate its capacitance!

July 29, 2026Physics

ScanSolve AI Answer

Answer

0.0111 \text{ F}

4 steps

Identify the given values.
The energy stored in the capacitor ( $E$ ) is $20.0 \text{ J}$ .
The voltage across the capacitor ( $V$ ) is $60.0 \text{ V}$ .
Recall the formula for the energy stored in a capacitor.
The energy stored in a capacitor is given by the formula:
$E = \frac{1}{2} C V^2$
where $C$ is the capacitance and $V$ is the voltage.
Rearrange the formula to solve for capacitance ( $C$ ).
To find $C$ , we can rearrange the formula:
$2E = C V^2$ $C = \frac{2E}{V^2}$
Substitute the given values and calculate the capacitance.
$C = \frac{2 \times 20.0 \text{ J}}{(60.0 \text{ V})^2}$ $C = \frac{40.0 \text{ J}}{3600 \text{ V}^2}$ $C = \frac{1}{90} \text{ F}$ $C \approx 0.0111 \text{ F}$
The capacitance of the capacitor is approximately 0.0111 F.
$0.0111 \text{ F}$